This same problem would happen in RN black holes as well. In any case, the strong cosmic censorship conjecture would ensure that the inner horizon is a singular surface. But I don't understand why the maximal extension would be a problem for the singularity theorem.

@noir1993 As explained, the Singularity Theorem states that all timelike and lightlike geodesics inside the horizon must end in a singularity. However, in a Kerr black hole they don’t, as shown by the wavy line on the diagram. Therefore the Singularity Theorem does not work for the Kerr black hole and by itself cannot ensure the existence of the singularity in spacetimes with multiple regions. The inner horizon is not a physical singularity and cosmic Censorship is a not a part of the Singularity Theorem.

But that is well known and is exactly why we need the two cosmic censorship conjectures. The weak one for the event horizon and the strong one for BHs with a Cauchy horizon. For example see Witten's lecture notes (section 6.1 page 60.) arxiv.org/abs/1901.03928

@noir1993 Sorry, but I don’t understand what cosmic censorship has to do with your original question. The question is why Roy Kerr thinks that black holes may not have singularities and the answer apparently is, because the Singularity Theorem cannot require a singularity in a rotating black hole. However, your Witten link is about naked singularities and how we can prohibit them by additional arbitrary restrictions outside of the gravitational equations. It is unclear how any of this is relevant to your original question.

As far as I can understand, you're saying that Kerr has an issue with the fact that we can extend certain geodesics across the inner horizon that end up in another universe. Strong cosmic censorship prevents that from happening.

@noir1993 Cosmic censorship is a conjecture suggested in addition to what follows from GR equations. Such conjectures may not contradict GR. When they do, they are considered disproven and incorrect. The Kerr solution is an exact solution of GR, so no add-on conjectures are allowed to contradict this solution. Specifically, cosmic censorship in its strongest version was disproven in 2018 by Mihalis Dafermos and Jonathan Luk for the Cauchy horizon of an uncharged, rotating (Kerr) black hole: quantamagazine.org/…

Dafermos showed that Penrose's original version of SCC doesn't hold, however there is a weaker version of SCC that holds for asymptotically flat black holes. Classically the only known example where SCC is violated is the RN-dS solution (although quantum effects can restore SCC). The fact that SCC holds for Kerr ensures that we can't extend the metric across the inner horizon with locally square integrable Christoffel symbols. Here's the story in Dafermos' own words web.math.princeton.edu/~dafermos/research/…

The bottom line is that Christodoulou's version of SCC holds for flat BHs and even if you can draw a wavy line across the Cauchy horizon, it's not important. Theres still a singularity at the Cauchy horizon that will not let you extend the spacetime (with locally square integrable Christoffel symbols). So it doesn't really matter if the geodesics that cross the Cauchy horizon end up at the singularity or not.

@safesphere In mathematics, conjectures may not contradict the axioms. But in physics, we have no axioms: experiment is the fundamental authority that gives credibility to our hypotheses, or falsifies them. No mathematical proof has such authority in physics.

@noir1993 Be that as it may, it was not a part of your original question. Based on your comments, your actual question involves Cosmic Censorship that is not even mentioned in the original question. Perhaps you should add an edit to the original question to clarify what exactly you are interested in the answer. This would make it easier for responders to satisfy your bounty.

@JohnDoty Your general point is not applicable here. Conjectures in GR are a part of GR and are not intended to contradict GR solutions. Let me use this parallel. I have a square coffee table of $1\,m^2$. The math gives me two solutions for the sides, $+1\,m$ and $-1\,m$. Then I use a conjecture that sides must be positive. This conjecture doesn’t contradict the solution, just removes the unphysical part. However, if my conjecture is, say, “the sides must be less that a square root of the area”, then this conjecture contradicts the math and must be ruled out. So the point in my comment stands.

@noir1993 BTW personally I agree with you that the Kerr solution beyond the inner horizon is unphysical. My only point is that you probably should add more information to your original question to make it more clear.

@safesphere You're asserting that GR is not a scientific discipline.

@safesphere How do you know that one solution is physical, but the other is not?

@JohnDoty Sorry, but I have no idea what exactly you are referring to. You should be more specific with your statements and questions. Then I’d be happy to respond. Also, please feel free to post your own answer to express your views and opinions, as this site is not for extended conversations in comments.

It's not complicated. Physics is an experimental science. What's your experiment here?

@JohnDoty I’d love to discuss if modern physics is even science anymore, but comments are not for discussions. If you have anything tangible to contribute to this particular Q&A, please post your own answer and I’ll upvote it, I promise.

"In a nutshell, the Penrose Singularity Theorem states that light inside the horizon cannot escape to infinity and therefore must end up in a singularity." This is not what the theorem states. The statement is that there are light rays with a finite affine length that cannot be extended. The example you give is a curve that actually has infinite affine length.

@TimRias "This is not what the theorem states" - Only if tomayto is different from tomahto. Can't you see the forest behind the trees? "The example you give is a curve that actually has infinite affine length" - Yes. If geodesics are not required to hit a singularity, then a singularity is not required to exist. BTW in some spacetimes geodesics with the infinite affine length still hit a singularity.

The singularity theorem does not state that all geodesics need to hit the singularity. It states that there exist null geodesics with finite affine length that cannot be extended.